如何將namevaluepairs中的MySQL數據庫連接數據從Android傳遞到PHP

Question

我正嘗試使用PHP連接到MySQL，從Android傳遞數據庫連接參數。我不想對連接參數進行硬編碼，也不想將它們存儲在單獨的文件中。當我在PHP中有數據庫參數時，我的代碼工作正常，但現在不能工作，因爲我試圖通過名稱值對將它們從Java傳遞到PHP，如下所示。

除了使用傳遞變量而不是硬編碼的PHP連接外沒有任何變化，所以我懷疑有些格式化或REGEX問題，但找不到任何解決方案。任何幫助非常感謝！

根據VolkerK的協助解決問題。查看原始的PHP代碼並在下面更新。

ORIGINAL SQLQuery.php：

<?php 
mysql_connect($_REQUEST['url'],$_REQUEST['username'],$_REQUEST['password']); 
mysql_select_db($_REQUEST['database']); 
$q=mysql_query($_REQUEST['SQL']); 
while($e=mysql_fetch_assoc($q)) 
     $output[]=$e; 
print(json_encode($output)); 
mysql_close(); 
?> 

工作SQLQuery.php：

<?php 
if (get_magic_quotes_gpc()) { 
    $process = array(&$_GET, &$_POST, &$_COOKIE, &$_REQUEST); 
    while (list($key, $val) = each($process)) { 
     foreach ($val as $k => $v) { 
      unset($process[$key][$k]); 
      if (is_array($v)) { 
       $process[$key][stripslashes($k)] = $v; 
       $process[] = &$process[$key][stripslashes($k)]; 
      } else { 
       $process[$key][stripslashes($k)] = stripslashes($v); 
      } 
     } 
    } 
     unset($process); 
} 

define('DEBUGLOG', true); 
$output = array(); 

$mysql = mysql_connect($_REQUEST['url'],$_REQUEST['username'],$_REQUEST['password']); 
if (!$mysql) { 
    $output['status']='Error'; 
    $output['errormsg']='MySQL connect error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array( 
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'username'=>$_REQUEST['username'], 
      'password'=>$_REQUEST['password'] 
     ); 
    } 
} 
else if (!mysql_select_db($_REQUEST['database'])) { 
    $output['status']='Error'; 
    $output['errormsg']='Database select error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array( 
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'database'=>$_REQUEST['database'] 
     ); 
    } 
} 
else if (false===($q=mysql_query($_REQUEST['SQL']))) { 
    $output['status']='Error'; 
    $output['errormsg']='Query error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array( 
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'SQL'=>$_REQUEST['SQL'] 
     ); 
    } 
} 
else { 
    while($e=mysql_fetch_assoc($q)) { 
     $output[]=$e; 
    } 
} 

print(json_encode($output)); 

從我的Android代碼（！細節改變，以保護無辜）提取物：

String phpDBURL = "mysqlserver.blah.com:3306"; 
String phpURL = "http://www.blah.com/php/"; 
String dbname ="dbref_Evaluate"; 
String username = "dbref_admin"; 
String password = "password"; 
String SQL = "SELECT ID, ShortDesc FROM User WHERE Account = 'myname@gmail.com'"; 
//the query to send 
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(); 
nameValuePairs.add(new BasicNameValuePair("url",phpDBURL)); 
nameValuePairs.add(new BasicNameValuePair("username",username)); 
nameValuePairs.add(new BasicNameValuePair("password",password)); 
nameValuePairs.add(new BasicNameValuePair("database",dbname)); 
nameValuePairs.add(new BasicNameValuePair("SQL",SQL)); 
Log.v("Common.SQLQuery", "Passing parameters: " + nameValuePairs.toString()); 
//http post 
try{ 
     HttpClient httpclient = new DefaultHttpClient(); 
     HttpPost httppost = new HttpPost(phpURL + "SQLQuery.php"); 
     httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 
     HttpResponse response = httpclient.execute(httppost); 
     HttpEntity entity = response.getEntity(); 
     InputStream is = entity.getContent(); 
//convert response to string 

等等。

Answer 1

您需要 更多 錯誤處理。 任何mysql_ *函數都可能失敗，您的代碼必須對此作出反應。

原油例如：

<?php 
define('DEBUGLOG', true); 
$output = array(); 

$mysql = mysql_connect($_REQUEST['url'],$_REQUEST['username'],$_REQUEST['password']); 
if (!$mysql) { 
    $output['status']='error'; 
    $output['errormsg']='database error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array(
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'username'=>$_REQUEST['username'], 
      'password'=>$_REQUEST['password'] 
     ); 
    } 
} 
else if (!mysql_select_db($_REQUEST['database'])) { 
    $output['status']='error'; 
    $output['errormsg']='database error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array(
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'database'=>$_REQUEST['database'] 
     ); 
    } 
} 
else if (false===($q=mysql_query($_REQUEST['SQL']))) { 
    $output['status']='error'; 
    $output['errormsg']='database error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array(
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'SQL'=>$_REQUEST['SQL'] 
     ); 
    } 
} 
else { 
    while($e=mysql_fetch_assoc($q)) { 
     $output[]=$e; 
    } 
} 

print(json_encode($output)); 

Answer 2

使用POST而不是PHP文件中的REQUEST